How do you Calculate the Δx for each of the following?

a.) an electron with Δv= 0.100 m/s
b.) a baseball with mass 145 g with Δv = 0.100 m/s

1 Answer
Aug 5, 2016

The version of the Heisenberg Uncertainty Principle as defined with uncertainties Deltavecx and Deltavecp_x is:

\mathbf(DeltavecxDeltavecp_x >= h)

Recall from physics that momentum is vecp_x = mvecv_x for mass m and velocity vecv, and that Planck's constant is h = 6.626xx10^(-34) "J"cdot"s", or "kg"cdot"m"^2"/s".

So, if you want to calculate the uncertainty in position:

DeltavecxmDeltavecv_x >= h

color(green)(Deltavecx >= h/(mDeltavecv_x))

You could resolve the inequality by simply solving for the minimum uncertainty of the x position. Then you would get:

  1. color(blue)(Deltavecx) = h/(m_eDeltavecv_x) = (6.626xx10^(-34) "kg"cdot"m"^2"/s")/((9.109xx10^(-31) "kg")("0.100 m/s")) = color(blue)(7.27xx10^(-3) "m")

  2. color(blue)(Deltavecx) = h/(m_"ball"Deltavecv_x) = (6.626xx10^(-34) "kg"cdot"m"^2"/s")/(("0.145 kg")("0.100 m/s")) = color(blue)(4.57xx10^(-32) "m")

The physical interpretation is...

  • You know to an OK certainty as to what the position of the baseball is. The radius of an average baseball is about 7.4xx10^(-2) "m", so your uncertainty is less than pm 10%.
  • For an electron, the uncertainty in velocity is quite low (an electron normally moves very quickly, and "0.100 m/s" is sluggish for an electron), and thus, the uncertainty for its position is quite high.

Think about the radius of an electron, which is somewhere around 10^(-16) "m". That means 7.27xx10^(-3) "m" is approximately 10^13 times as long as the radius of an electron. That's a horrendous uncertainty!